The area of the region bounded by the lines $\frac{x}{7 \sqrt{3} a}+\frac{y}{b}=4, x=0$ and $y=0$ is:

$56 \sqrt{3} ab$ $56 a$ $\frac{a b}{2}$ $3 ab$

$56 \sqrt{3} ab$

The correct answer is Option (1) → $56 \sqrt{3} ab$